unit 4 equilibria including acid base answers
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Unit 4 Equilibria Including Acid Base AnswersTRANSCRIPT
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5/19/2018 Unit 4 Equilibria Including Acid Base Answers
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1. (a) (i) 2Ca(NO3)2 2CaO + 4NO2 + O2formulae correct (1)balance (1). Ignore any state symbols.
The balance mark is notstandalone. 2
(ii) steam / fizzing sound / crumbles (1)
solid swells up / milky liquid produced / comment about sparingly soluble
substance (1)
CaO + H2O Ca(OH)2 (1) ignore any state symbols 3
(iii) less (1) 1
(iv) (Cat)ion size increases down the Group / charge density decreases (1)
(not atom size)
The polarizing power of the cation decreases down the Group (1).
The less polarized the anion is by the cation the more difficult the nitrate is to
decompose (1).
Polarisation mark could come from the less the electron cloud is distorted
or
trend in cation size (1)
comparison of the lattice energies of the nitrate and the oxide (1)
balance in favour of oxide at top of group
and the nitrate at the bottom (1) 3
(b) (i) same number of particles in a smaller volume / gas density increased (1) 1
(ii) comment related to the number of molecules on each side to explain a shift to
l.h.s. (1) (not just due to Le Chatelier)
so at higher pressure equilibrium moves to favour N2O4 (1) 2
(iii) Kp = p(NO2)2
There must be some symbolism for pressure, and no [ ] 1
p(N2O4)
(iv) (Kp =p(NO2)2
p(NO2)2 = 48 0.15 = 7.2 (1)
p(NO2) = 2.7 (1) atm (1) accept 2.683 / 2.68 / 2.7
= 48)
p(N2O4)
Answer and units conditional on (iii). 3[16]
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2. (a) (i) HCl: pH = 1.13
[H+] = 0.074 mol dm3
[HCl] = 0.074 mol dm3 [0.074 to 0.07413] 1
(ii) HOCl: pH = 4.23
[H+] = 5.89 105mol dm3 (1)
Ka= [H+] [OCl]
[H+] = [OCl](1) or implied later in calculation
[HOCl] = [H+]2/ Ka= 0.0932 mol dm3 (1) 4
(1)
[HOCl]
(b) (i) [H+] = 0.10 / 0.1047 / 0.105(1)
(ii) H2SO4 H+ + HSO4
(1) or
H2SO4+ H2O H3+O + HSO4 ignore state symbols
HSO4 H+ + SO4
2(1) Must be
H2SO4+ H2O H3O++ SO4
2 ignore state symbols
(iii) second ionisation suppressed by the first ionisation (1) 4
(c) (i)][O[HCl]
O][H][ClK
24
22
22
c
= 1
(ii) 4HCl + O2 2Cl2 + 2H2Oequilibrium mols 0.20 0.050 (1) 0.30 and 0.30 (1)
[ ] eq 10(1) 0.020 0.0050 0.030 0.030
Kc= [0.030]2x [0.030]2 = 1010 or 1012 or 1013 or 1012.5
[0.020]4x [0.005] (mol1dm3)(1) 4
(d) (i) As reaction (left to right) is exothermic (1)
Decrease in temperature drives equilibrium to from left to right(1) 2
(ii) As more (gas) molecules on the left (1), equilibrium is driven from left to right(1)
2
(iii) A catalyst has no effect(1)
As it only alters the rate of the reaction not the position of equilibrium / it alters the
rate of the forward and reverse reactions equally (1) 2[20]
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3. (a) Still reacting/ rate of forward reaction and backward reaction equal /implication that
forward and backward reactions are still taking place(1)
But concentrations constant / no macroscopic changes (1) 2
(b) Temp (Increases) Left / to SO2/ to endothermic / lower yield (1)
Press Increases/faster (1) Right to SO3/ to smaller number of molecules (1)
3
(c) (i) Increases rate / or suitable comment on rate (1)
Moves position of equilibrium to endothermic side / or suitable comment on
equilibrium such as reasonable yield / less SO3(1)
Either compromise in which the rate is more important than the position of
equilibrium
oroptimum temperature for catalyst to operate
or
valid economic argument (1) 3
(ii) Increases rate / more SO3/only needs small pressure to ensure gas passes through
plant / high or reasonable yield obtained at 1 atms or at low pressure anyway(1)
and
references to economic cost against yield benefit
e.g increase in pressure would increase yield of product but the increase in yield
would not offset the cost of increasing the pressure (1) 2
(iii) Catalyst speeds up reaction(1) 1
(d) Vanadium (V) oxide / vanadium pentoxide / V2O5(1) 1
(e) Any one use
production of fertilizers, detergents, dyes, paints, pharmaceuticals (in) car batteries,
pickling metal 1[13]
4. (a) only partially dissociated / ionised / not fully dissociated(1)
into H+ions / H3+O / proton donor(1) 2
(b) Ka =[HA]
]][AO[H 3+
(1) 1
(c) (i) 9.0 to 9.4(1) 1
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(ii) 9.0 to 9.4(1)
or same answer as (c)(ii) 1
(iii) pKa= 5.6(1)
Ka= 2.5 106(1)consequential 2
(d) (i) (a solution that) resists change in pH / retains an almost
constantpH(1)
on addition of smallquantities of acid or alkali(1) 2
(ii) 5.2 to 5.8(1)
5.5 or 5.6(1)or answer from (c) (iii) based on misreading scale
of graph, eg. 4.8 2
(e) Phenolphthalein(1)
indictor changes colour between pH 7 and 10 this is vertical part of graph(1)
methyl orange would change in acid / give pH between pH4 and pH6(1)
n. b. must be +ve statement about methyl orange 3
(f) exothermic reaction / heat (energy) released during reaction(1)
HCl is strong acid fully ionised(1)
this is weak acid so some energy used for dissociation (1) 3
(g) (i) pH = log(10) [H+] or in words 1
(ii) 1.8 105=1
][H 2+(1)
[H+] = 1.8 105= 4.24 103(1)
pH = log (4.24 103)= 2.37/2.4(1)2 to 4 sig. figs. 3
[21]
5. (a) (i) Kp = 2SO3
O2
2
SO2
P
PP (1)
[ ] no mark
( ) OK 1
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(ii) 2SO3 2SO2 + O2
Mols at start 2 0 0
mols at equ 0.5 1.5 0.75 (1)
Mark by process
1 mark for working out mole fraction1 mark for 10
1 mark for correct substitution in Kpand answer
1 mark for unit
i.e. PSO2=75.2
5.1 10 = 5.46
PO2=75.2
7.0 10 = 2.73
PSO3
=75.2
5.0 10 = 1.83
n.b. could show mole fraction for all 3 and then
Kp = (5.46)2 (2.73) / (1.83)2= 24.5(1)atm(1) 5
10 later to
give partial pressure.
(b) (i) No effect(1) 1
(ii) No effect(1) 1[8]
6. (a) (i) fraction of the total pressure generated by a gas or
or
pressure gas would generate if it alone occupied the
volume
or
Ptotal mol fraction(1) 1
(ii) O)p(H)p(CH
)p(Hp(CO)K
24
32
p
=
(1) not [ ] 1
(iii) Increase in total pressure will result in less product
molecules in the equilibrium mixture / equilibrium
moves to left(1)
because more molecules on product side of the equilibrium
than on left(1) 2
(b) (i) No change(1) 1
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(ii) KPincrease(1) 1
(iii) No change(1) 1
(c) (i))p(CH
1K4
p = (1) 1
(ii) 9.87 103kPa1/ 9.87 106Pa1consequential on (i)(1) 1
Allow 3 5 sig fig
(iii) equilibrium has moved left in favour of gas(1)
exothermic going left to right/in the forward direction / as
written(1)Stand alone 2
(iv) Answer yes or no with some sensible justification(1)
e.g. No the costs would not justify the amount produced 1[12]
7. (a) Few molecules dissociate (into protons) / partially dissociated / ionised(1) 1
Not fully dissociated scores zero
(b) Maintains an almostconstant pH / resists change in pH(1)
with the addition of smallamounts of acid or alkali(1) 2
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(c)
14
12
10
8
6
4
2
0 10 20 30 40 50
pH
Volume of alkali added / cm3
starting pH(1)at 2.8
endpoint(1)vertical between 6 and 11 including 7-10
vertical(1)at 25 cm3
general shape(1)finish above 12 4
(d) Almost horizontal area marked on graph(1) 1
(e) (i)COOH][CH
]][HCOO[CH
3
3a
+
=K (1)
or
COOH][CH
]O][HCOO[CH
3
33a
+
=K (1) 1
(ii) pH = pKaat half way to neutralisation point = 12.5 cm3(1)
This could be shown on the graph
because pH = pKawhen [CH3COO] = [CH3COOH](1) 2
[11]
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8. (a) The marks are for:
writing the expression for K
substituting correctly
calculating p(SO3)
correct generation of the ratio
calculation of the ratio to give answer which rounds to 95 t
Kp = pSO32/ pSO2
2pO2(= 3.00 104)(1)
3.00 104= pSO32/ 0.1 0.1 0.5(1)if no expression for Kp is
given this correct substitution can score 2 marks
pSO32= 150
pSO3= 12.25(1)Ratio of SO3= 0.5)0.1(12.25
100%12.25++
(1)= 95%(1) 5
(b) (i) The marks are for
Recognizing the existence of hydrogen bonds ( between molecules)(1)
That each molecule can form more than one hydrogen bondbecause of the two OH (and two S=O groups) / or a description of
hydrogen bonds in this case / or a diagram showing the
hydrogen bonds (1)
That hydrogen bonds make for strong intermolecular forces(and hence high boiling temperature) which requires higher
energyto break / separate molecules (1) 3
(ii) If water is added to acid heat generated boils and liquid spits out(1)
if acid added to water the large volume of water absorbs the heat
generated (and the mixture does not boil)(1) 2
(c) (i) pH = log10(0.200) = 0.70(1)
allow 0.7 or 0.699 1
(ii) realising that the firstionisation / dissociation of sulphuric
and that of HCl are bothcomplete(1)
that the second ionisation of sulphuric is suppressed by the
H+from the first(1)
little contribution from 2nd ionisation so reduces the pH
very little / increases the [H+] very little(1) 3
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(d) (i) Lead equations 1 mark
Pb + H2SO4PbSO4+ 2H++ 2e(1)
or
Pb + SO42 PbSO4+ 2e
Lead(IV) oxide equations 2 marks
PbO2+ H2SO4+ 2H++ 2ePbSO4+ 2H2O
or
PbO2+SO42+ 4H++ 2ePbSO4+ 2H2O+
Species(1)balancing(1) 3
(ii) PbO2+ Pb + 2H2SO42PbSO4+ 2H2O (1) 1
[18]
9. (a) (i) pH = log10[H+] / pH = lg [H+](1) 1
(ii) KW= [H+] [OH] or KW= [H3
+O] [OH](1) 1
(b) fully ionised / fully dissociated / almost completely ionised(1) 1
(c) (i) 0.70 (or 0.699)(1) 1
(ii) [H+] = KW/ [OH] = 1.25 1014(1)
pH = 13.9 or 13.90(1) 2
(d) (i) Ka= [HA]
]][A[H +
(1) 1
allow [H3+O]
(ii) [H+] = (Ka[HA])(1)= 0.00474(1)
pH = 2.32 / 2.33(1) 3
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(e) [H+] =][A
[HA]K
a(1)
[H+] = (5.62 1050.3) / 0.6 = 0.0000281 / 2.81 105(1)pH = 4.55(1) 3
Or
pH = pKa+ log [HA]
][A
= log10(5.62 105) + log10 ]300.0[
]600.0[= 4.55
If initial error in statement of [H+] or Henderson equation max 1[13]
10. (a) (i) Kc= [SO3]2/ [SO2]
2[O2](1) 1
(ii)30
2.0
60
1.0
60
8.1
= 3.33 103 1.67 103 0.03(1)
Kc= 323
2
1067.1)1033.3(
)03.0(
= 4860or 4.86 104(1)
mol1dm3(1) 3
(b) (i) Kcdecreases(1) 1
(ii) shifts to left / in reverse(1) 1
(c) (i) no effect(1) 1
(ii) no effect(1) 1
(d) (i) Kp= pSO32/ pSO2
2pO2(1)penalise square brackets 1
(ii) Total number of moles(1)consequential on a (ii)
SO2= 0.0952(4); O2= 0.0476(2); SO3= 0.857 (1)(1) 2
(iii) Partial pressures: SO2= 0. 190 (5) atm; O2= 0.0952 (4) atm; 1
SO3= 1.71(4) atm(1)i.e. multiply answer in (ii) by 2
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(iv) 1.7142/ 0.190520.09524 = 850(1)
atm1(1) 2[14]
11. (a) (i) Gases have much higher entropies than solids as there are many more
ways of arranging the entities / less ordered / more random(ness)
OR reverse argument(1)
ZnCO3has more atoms/is more complex than ZnO (1) 2
(ii) Ssystem= (+43.6) + (+213.6) (+82.4)
=+174.8/175 J mol1K1
method(1)answer, sign and units (1)
Correct answer, sign and units with no working (2) 2
(b)
As printed
Ssurroundings=T
H
OR = 298
)105.464( 3+
(1)
= 1560 / 1559 J mol1K1
answer, sign and units (1)
Amended
Ssurroundings=T
H
OR = 298
)100.71( 3+
(1)
= 238(.3) J mol1K1
answer, sign and units (1)
ONLY accept 3 or 4 SF 2
IF correct answer, sign and units with no working (1)
(c) (i)
Stotal= +174.8 1558.7
= 1384 / 1380 J mol1K
1
IF + 174.8 1560
= 1385(.2)
= 1385 / 1390 J mol1K1
IF + 174.8 1559
= 1384 J mol1K1
= 63.5 / 64 / 63 / 63.2 / 63.4 J mol1K1
ONLY penalise incorrect units OR no units in (a)(ii), (b) and (c)(i) once 1
(ii) Natural direction is right to left /reverse as Stotal/total entropy change
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is negative / less than zero. 1
MUST be consistent with (i)
(d) (i) Kp=p co2((g) eqm) 1
(ii) Increase temperature / reduce pressure (1)
Decreases Ssurroundings(negative) and hence increases Stotal/ Le Chateliers
principle applied (i.e increasing temperature, reducing pressure) (1) 2[11]
12. (a) (i) Pairs up CH3CH2COOH and CH3CH2COO
and H2O/H3O+(1); correct identification of which is acid and
which base (1) 2
(ii) Ka= [CH3CH2COO] [H3O
+] / [CH3CH2COOH] (1) 1
[H+] is acceptable.
(iii) [H+] = (Ka[HA])or Ka[HA] (1)= (1. 3 105 0.10)
= 1.14 103 mol dm3(1)pH = 2.9 or 2.94, i.e. to 1 or 2 d.p. (1) 3
Consequential on the value of [H+] provided the pH resulting is
between 0 and 7.
(iv) [H+][OH] = 1014 (1) = 1.14 103[OH]
Thus [OH] =1014 103(1)
= 8.77 (8.8) 1012mol dm3(1)units needed (2 or 3 sf)
Consequential on the answer to (iii) for [H+]
Allow 8.71 1012if solved using pH + pOH and pH = 2.94;
7.9 1012if solved using pH + pOH and pH = 2.9. 3
(b) CH3CH2COO+ H2O CH3CH2COOH + OH (1)
Hydroxide ions make the solution alkaline (1)or propanoate
ion deprotonates the water
or CH3CH2COONa + H2O CH3CH2COOH + NaOH (1)
Explanation then must comment that acid is weak/not fully
ionised 2
(c) (i) Solution that maintains almost constant pH (1)
for small addition of acid or alkali (1) 2
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(ii) pH = pKa + lg [salt]/[acid] (1)
= 4.9 + Ig (0.05)/(0.025) (1)for dividing by 2
= 5.19 or 5.2 (1). 3
If the Henderson equation is wrong but concs are divided by 2
then 1/3 max.Or
[ ] )1([salt]
Ka[acid]H =+
)1(0.050
025.0101.30 -5 =
pH = 5.19 or 5.2 (1)
If the concns are twice what they should be, ie. candidate does
not spot the volume increase, then max (2). The pH is still 5.2,
so care is needed.[16]
13. (a) pH = Ig [H+(aq)]
OR [H+(aq)] = 10pH/ 109.6= 109.6(method) (1)
= 2.5(1) 1010mol dm 3(2) 3
(b) (i) Kw= [H+(aq)][OH(aq)] 1
(ii) [OH(aq)] =( )[ ]aqH+wK
=10
14
1051.2
100.1
= 3.98 / 4(.0) 105mol dm3 1
(iii) [Ca(OH)2(aq)] = 0.5 3.98 105)
= 1.99 / 2(.0) 105mol dm3 1
(iv) =1.99 / 2(.03) 105 74
=0.00147 (gdm3) MUST be to 3 SF 1
(v) Due to reaction with carbon dioxide in the air / temperature differences 1
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(c) (i) pH = 3(.0) Penalise if more than 2 SF 1
(ii) Amount of Ca(OH)2=1000
1099.1100 5
= 1.99 106mol (1)
Amount of HCl =1000
1099.11002 5
= 3.98 106mol (1)
Volume of HCl =1000
1099.110021000 5
= 4(.0) / 3.98 cm3 OR 4(.0) / 3.98 103dm3(1) 3
(iii)
Verticalportion of the graph (between pH 7 and 4)at about 4 cm3(1)
Correct initial and finishing pH (910 and 3) and general shape (1) 2
(iv) Phenolphthalein changes colour outside pH range of end point/ pH range
of phenolphthalein is too high. 1[15]
14. (a) (i) NH3base and NH4+acid (1)
H2O acid and OHbase (1)
OR
linking (1)
acid and base correctly identified (1) 2
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(ii) Starting pH of (just above) 11 (1)
Graph showing vertical line between pH 4 and 6
With vertical section 35 units in length (1)
at a volume of HCl of 20 cm3(1)
Final pH of between 1 and 2 (1) 4
(iii) Named indicator consequential on vertical part of their graph (1)
Because allof its range is withinthe vertical part of the graph /
pKind 1 is within vertical part of graph / it changes colour
completely/ stated colour change (MO: yellow red; BB: blue
yellow; PP: pink colourless) within the pH of the vertical part of
the graph (1) 2
(b) (i) Ka= [ ][ ]2
23
HNO
NOOH +
square brackets essential 1
(ii) [H+] = [NO2] or [H+]2= Ka [HNO2] (1)
[H+] = (Ka 0.12) = 0.00751 mol dm3(1)
pH = log [H+] = 2.12/2.1 (1)
ALLOW any correct conversion of [H+l into pH provided the
answer is less than 7 3
(iii) Moles NaNO2= 1.38/69 = 0.020 (1)
[NO2] = 0.020 / 0.10 = 0.20 (mol dm3)
[H+] =salt][
acid][aK =20.0
120.01070.4 4 = 2.82 10 4(1)
pH = log 2.82 104= 3.55 /3.6 /3.5 (1) 4
(iv) In a buffer both [acid] and [salt] must be large compared to the
added H+or OH-ions (1)
but in NaNO2alone [ HNO
2] is very small (1)
OR
to remove both H+and OHthere must be a large reservoir of
both NO2ions and HNO2molecules (1)
which there are a solution of NaNO2and HNO2but not in NaNOalone (1) 2
[18]
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15. (a) PressureNOTpartial pressure )
intensity or change of colour ) Any one
volume ) 1
(b) Kc= [NO2(g)]2/ [N2O4(g)]
State symbols required 1
(c) Mol NO2at equilibrium = 0.0120 / 1.20 102(1)
Kc= (0.0120)2 (0.0310)
= 4.6 / 4.65 103(1)mol dm3(1) 3
(d) (i) Amount of NO2reduced 1
(ii) No effect 1
(e) As Kcis bigger, more NO2is produced so heat helps forward reaction /
by Le Chateliers principle reaction goes forward to use up heat /
as temperature increases Stotalmust be more positive so Ssurroundings(=H/T must be less negative 1
(f) Positive / + with some attempt at explanation(1)
1 mol / molecule gas 2 mol / molecule gas / products more disordered
than reactants (1) 2
(g) Ssurroundings= T
H OR
T
H 1000 1
(h) Stotalis positive as reaction occurs (1)
So Ssystemmust be more positive than Ssurroundingsis negative (1) 2
[13]
16. (a) (i) pH = log(10)[H3O+] (1)
OR [H+]instead of [H3O+] applied throughout 1
(ii) [H3O+] greater than 1 (mol dm 3) 1
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(b) (i) Ka. =[HA]
][A]O[H 3+
(1)
[H3O+] = 1.14 103mol dm 3(1)
pH = 2.9(4) (1) 3
(ii) Start at the same pH as in (i) (1)
Graph showing vertical at 25 cm3(1)
vertical section 35 units in length with midpoint around pH 8 (1)
general shape correct including buffer zone and final pH not > 13 (1) 4
(iii) pH = pKa 1
[10]
17. (a) Pressure exerted by the gas if it alone occupied the same volume at the
same temperature/mole fraction total pressure 1
(b) (i) Kp= 222
)NO(
)O()N(
p
pp 1
(ii) Correct number of moles (1)
Correct mole fractions (1)
Correct partial pressures (1)
2.45 103(1)ACCEPT 24 SF 4
(c) Kpincreases (1)
Equilibrium moves to r.h.s. (1)which is the exothermic direction (1) 3
(d) (i) Kp= p(Ni(CO)4) /p(CO)4 1
(ii) High partial pressure with some reason (1)
so the pressure Ni(CO)4increases to keep Kp constant. (1) 2
[12]
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18. (a) (i) Methanol is the biggest/ most complex molecule / greatest MR/most
atoms/most electrons 1
(ii) Ssystem= 239.7 197.6 2(130.6)
= 219.1/ 219 J mollK1
Method (1)answer + units (1) 2
(iii) yes as 3 molecules 1ORyes as (2) gases a liquid 1
(iv) Ssurr= H/T (stated or used) (1)
= (129/ 298) = +0.433 kJ mol1K1/ +433 J mol1K1/+ 432.9 (1)
1 for wrong units/ no units / more than 4 SF
1 for wrong sign/ no sign 2
(v) Stotal= 219.1 + 433 = +213.9 / +213.8 J mol1K1/ +214 J mol1K1/
+0.214 kJ mol 1K1(1)
Positiveso possible(1) 2
(b) (i)
Faster at 400C (1)
even though yield is lower (1)
Temperature
Pressure
(ii) Not in same phase as reactants. ALLOW stateinstead ofphase 1
Higher pressure improves yield of methanol (1)
Higher pressure increases rate (1)
Maximum 3 3
(iii) Kp=p(CH3OH)/p(CO)p(H2)2 1
(iv) Partial pressure of methanol = 200 55 20 = 125 atm (1)
Kp= (125)/55202
= 5.68 103/ 5.7 103atm2(1) 2
(c) (i) Number of molecules / fraction of molecules with energy EA/number
of molecules which have enough energy to react. 1
(ii) Vertical line / mark on axis to show value to the left of lineEA 1
[17]
19. (i) Weak acid is dissociated to a small extent/slightly dissociated/
ionised/few molecules dissociate
ALLOWpartial dissociation
NOTnot fully dissociated. 1
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(ii) Ka =COOH]H[C
]COOH][C[H
94
94
+
OR [H3O+] for [H+] 1
(iii) Ka = [H3O+]2/ [acid] OR[H3O
+] = Ka [acid](1)
[H3O+
] = 1.23 103
(1) dependent on 1st
markpH = 2.91/2.92(1)ACCEPT2.9 1 or 2 d.p.
Correct answer with working(3)
Correct answer with no working(1)
ALLOW TE only if pH below 7 3
(iv) starting pH 2.9ALLOW starting in 2ndor 3rdboxes above pH 2(1)
consequential on (iii)
pH range vertical max 6 to 12 min 710(1)
Equivalence point at 25cm3(1)
General shape of curve andfinish at pH between 1213(1)
and end in 1stthree boxes above 12, extending to 4050 cm3
If drawn wrong way round2 maxie equivalence point(1)and
vertical drop (1)marks can be awarded 4
(v) Thymol blue(1) Consequential on (iv)
(Completely) changes colour within vertical portion/the working
range of the indicator is within the vertical portion / pKind 1 in
vertical position / pKindin centre of vertical position(1) 2
[11]
20. Penalise unitsonly once in this question
(a) (2192.3)[191.6 + (130.6 3)] (1)
= 198.8/199 J mol1K 1 (1) 2
(b)298
/2.92
298
/10002.92 H / T (1)
= + 309(.4) J mol1K1/ + 0.309(4) kJ mol1K1 (1) 2
(c) (i) 198.8 + 309 = + 110 J mol1K1(3 SF)
OR
198.8 + 309.4 = + 111 J mol1K1(3 SF)
[Do not penalise missing + sign if penalised already in (b)]
NOT4SF. Penatise SF only once on paper 1
(ii) Yes, as Stotalis positive / total entropy change 1
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5/19/2018 Unit 4 Equilibria Including Acid Base Answers
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(d) (i) Higher T makes Ssurroundingsdecrease (so Stotalis less positive) 1
(ii) Cost (of energy) to provide compression/ cost of equipment
to withstand high P/ maintenance costs.
NOTsafety considerations alone 1
(iii) Different phase/state (to the reactants) 1[9]
21. (a) (i) KP= p (CO2) allow without brackets, IGNOREp [ ] 1
(ii) 1.48 (atm)
Penalise wrong unit
Answer is consequential on (a) (i) e.g.48.1
1must have atm1 1
(b) (i) Kp= 2
22
(p(NOCl))
p(NO))p(Cl allow without brackets, penalise [ ] 1
(ii) 2NOCl 2NO + Cl2
Start 1 0 0
0.22 +0.22 +0.11eq moles 0.78 0.22 0.11 (1)
total moles of gas 1.11
mole fractions above values 1.11 (1)
0.7027 0.1982 0.09910
partial pressure / atm above values 5.00 (1)
3.51 0.991 0.495
Kp= 2
2
atm)51.3(
atm)(0.991atm495.0 (1)
= 0.0395/ 0.0394 atm (1)
range of answers 0.0408 / 0.041 0.039 / 0.0392 NOT 0.04
ACCEPT 2 S. F
Correct answer plus some recognisable working(5)
Marks are for processes
Equilibrium moles
Dividing by total moles
Multiplying by total pressure
Substituting equilibrium values into expression for KP 5
Calculating the value of KPwith correct consequential unit.
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(iii) As the reaction is endothermic stand alone(1)
the value of KPwill increase (as the temperature is increased) - (1)
consequential on 1stanswer (if exothermic (0) then KPdecreases(1))
For effect on KPmark, must have addressed whether reaction is
endothermic or exothermic 2
(iv) Because (as the value of KPgoes up), the value of
pCl2 (pNO)2/ (pNOCl)2(the quotient) must also go up (1)
and so the position of equilibrium moves to the right stand alone(1)
But mark consequentially on change in K in (iii)
If position of equilibrium moves to right so Kpincreases(max 1) 2
IGNORE references to Le Chateliers Principle[12]
22. (a) CH3COOH labelled as base and linked to CH3COOH2+labelled
(conjugate) acid (1)
H2SO4labelled acid and linked to HSO4labelled (conjugate) base (1)
If acids and bases correct but not clearly or correctly linked 1 (out of 2)
Just link but no identification of acids and bases (0) 2
(b) (i) (pH) more than 7 / 8-9 (1)
Indicator: phenolphthalein ALLOW thymolphthalein OR thymol blue(mark consequentially on pH)(1) 2
Mark consequentially on pH but if pH7 do not allow either methyl
orange or phenolphthalein
QWC*(ii) As OH/ base removes H+ions / Hneutis per mole of H2O produced / (1)
H++ OH= H2O
the equilibrium shifts to the right (1)
and so all the ethanoic acid reacts (not just 1% of it) (1)
OR
Endothermic (OH) bond breaking (1)
is compensated for (1)
by exothermic hydration of ions (1)
OR
H for CH3COOH + H2O CH3COO+ H3O
+= +2kJ mol1/ almost zero /
very small (1)
Hneut[CH3COOH] = +2 + Hneut[HCl] (1)
the same (for both acids) (1)OR
Hneutis per mole of H2O produced (1)
(heat) energy required for full dissociation (of weak acid) (1)
so Hneutslightly less exothermic (for weak acid) (1) 3
(iii) [H+]2= Ka[CH3COOH] = 1.74 105 0.140 = 2.44 106
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[H+] = 0.00156 (mol dm3) (1)
pH = 2.81 consequential on [H+] but not pH>7(1)
ACCEPT 2.80/2.8 (answers to 1 or 2 dp)
The assumptions are two from:
[H+] = [CH3COO] this mark can be earned from working /
negligible [H
+
] from ionisation of water (1)[CH3COOH] = 0.140 [H
+] 0.140 (mol dm3) / ionisation of acid
negligible (1)
solution at 25C (1) max 2 4
(iv) 1.74 105=][
][][
acid
saltH+(1)
[H+] = 1.74 105 0.070
[14]
= 1.22 105(1)
0.100
pH = 4.91 / 4.9 / 4.92NOT 5
Max 2 if 0.140 / 0.200 is used(1) 3
23. (a) (i) +313.4 4 197.6 29.9 (1)Absence of 4 (0)
= 506.9 J mol1K1OR507 J mol1K1ORanswerinkJ(1)
NOT 510 ie 2 SF
Missing or wrong units ie answer does not match units 1 max 2
(ii) Negative as expected because only 1 mole of gas on the RHS but 4
moles of gas on the LHS
Mark can be awarded if answer based on moles only rather than
states [5 moles 1 mole] 1
(iii) Ssurroundings = H/T (1)For equation or use of equation= 191 000 / 323
= (+) 591 J mol1K1(1)ORanswer in kJ 2
ALLOW (+) 591.3 J mol1
K1
/ (+) 590 J mol1
K1
ALLOW 2, 3 or 4 SF
+ sign not needed provided there is evidence in calculation to
show positive
In (i) and (ii) missing units is penalised once UNLESS a
different unit error is made hence penalise twice)
(iv) Stotalis positive/(+84 J mol1K1) so reaction should go forwards
ALLOWTEfrom(a)(i) 1
(b) (i) Kp= 44)(
pco
COpNiif square brackets [ ] are used (0) 1
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(ii) Ni(CO)4moles at eq 0.25 (1)
total number of moles = 99.25
PNi(CO)4=25.99
25.0 1 = (0.00252) PCO=
25.99
99= (0.9975) (1)
2nd
mark must be to at least 3 SF or working must be shown ie
some evidence that their total number of moles
Kp = 0.25 / 99.25 / (99/99.25)4
= 2.54 103(1)atm3(1)
Units marked independently
ALLOW 2.5 103
Many have total number of moles as 100 even when it is not
ALLOW TE for 2nd and 3rdmark if it should be 100 or just TE
for third mark if it shouldnt be 100 from their working
[Kp = 2.6 or 2.60 103] 4
(iii) Increasing the pressure/ concentration of CO would force the
reaction to the RHS with the smallest number of gaseous molecules
(1)
NOTpressurewithrate
Reduce the temperature so that the reaction goes in the exothermic
direction / increase the temperature to increase rate (1)
Do not allow equilibrium to be reached by passing the CO over the
nickel/recovering the product formed (1)
Use a catalyst to increase rate / increase the surface area of the nickel
to increase number of collisions (1) 3
(c) The reaction can be reversed by increasing the temperature (1)
as Ssurroundingswill become less positive/more negative as the temperatureis increased (and Ssystemwill remain almost unchanged) so Stotalbecomesnegative for the forward reaction (1) 2
[16]
24. (a) starts at 2.2 (1)vertical section at 40cm3of sodium hydroxide (1)
vertical section centred between pH 8-9 and between 2 to 3 squares high (1)
shape to include initial jump and finish between pH = 12 13 (1)
If curve drawn back to front, only 2ndand 3rdmarks available 4
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(b) (i) maintainsnearly constant pH / resists change in pH (1)
on adding small amountsof acid or alkali (1) 2
(ii) [ ] ][][
H salt
acidKa
=
+
OR pH = pKalg
[ ][salt]
acid
(1)
[H+] = 1.78 104125.0
25.0 (1)
[H+] = 3.56 104(mol dm3) 3
pH = 3.4(5) (1)
IGNORE no. of decimal places but penalise pH = 3
(iii) acid partially ionised and salt fully ionised
ORequations (1)
HA + OHA + H2O (1)
ALLOWH++ OHH2O followed by more dissociation of HA
A+ H+ HA (1)
[HA] and[A] are large (relative to H+and OHadded) / large reserves of
undissociated acid andsalt (and so the values of [HA] and [A] do not
change significantly) (1)
NOTE: If no equations given for effect of adding OH
and H
+
, correctexplanation can score (1)out of these two marks. 4[13]
25. (a) (i) Negative with some sensible explanation eg fewer moles of product (1)
3 moles of gases going to 2 moles of gases (1) 2
MUST mention gases or no changes in state
(ii) Positive with some explanationeg exothermic so surroundingsgain entropy (1)
Ssurroundings= T
H [OR given in words]
OR
Stotal= Ssystem+ Ssurroundings [OR given in words]as reaction goes, Stotalmust be positive therefore Ssurroundingsmust be positive
OR
Surroundings gain energy so more ways of arranging energy (1) 2
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5/19/2018 Unit 4 Equilibria Including Acid Base Answers
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(b) (i)
22
2
ONO
2
NO
PP
P)Kp(
= (1)
Check that it isnota + on denominator.
ALLOW ( ) but NOT [ ] eg ALLOW( 2
NO )(P 2 )2etc
ALLOW(pNO2)
2
Atm1/ Pa1/ kPa1/ m2N1 (1) 2ndmark dependent on 1st
ALLOWatms1/ atmospheres1
NOTatmetc
NOTKpa1 2
(ii) Temperature
but the lower temperature may slow the reaction down too muchOR reverse argument (1)
A lower temperature is needed to get a better yield (and would
cost less) because the reaction is exothermic (1)
Pressure
It will also increase the rate of the reaction (1)
A high pressure will increase yield as only two moles on the right
compared to three on the left/less moles on the right hand side (1)
Low pressure because of cost only gets mark if higher yield
at higher pressure identified
To award any of the yield marks must say why 4
(c) (i) Must be a quantity that can be measured
Eg
The pressure could be measured (1)
as it will decrease as the reaction proceeds because there are only
two/fewer moles on the right compared to three on the left (1)
OR colour (1)
as the nitrogen(IV) oxide is brown whereas the other gases
are colourless (1)
OR total volume (1)
which will decrease by one third/because there are fewer moles (1)
ALLOWacidity because NO2acidic andothers not (1 max)
NOTdilatometry
NOTtemperature 2
(ii) [NO] second order (1)
because when conc of NO is doubled, the rate goes up four times (1)
[O2] first order (1) 3
Then (iii), (iv) and (v) must follow consistently from (ii)
(iii) ALLOW TE from (ii) e.g.
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5/19/2018 Unit 4 Equilibria Including Acid Base Answers
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rate = k[NO]2[O2] rate = k[NO][O2] 1
(iv) third / 3 second / 2 1
(v) 8000 (1) dm
6
mol
2
s
1
(1) 8 (1) dm
3
mol
1
s
1
(1)Units can be given in any order 2
(d) The activation energy must be low
ORbond energies low
NOT more successful collisions
NOT large rate constant 1[20]
26. (a) (i) - lg (0.05) = 1.3(0) 1
IGNORE sig figs from this point on in this question
(ii) [OH] = 1 1014/ 0.05 = 2 1013(mol dm3) 1
OR via pOH
Correct answer with no working (1)
(b) (i) Ka= [H2PO4
][H3O+
]/[H3PO4]NOTusing H+instead of H3O
+ 1
(ii) [H3O+] = 101.20
= 0.063 (mol dm3)(1)
Ka=063.0500.0
063.0 2 (1) NOT consequential on (b)(i)
= 9.08 103(1)mol dm3(1)
= 9.11 103if [H3O+] not rounded
ALLOW
Ka=500.0
063.0 2(1)
= 7.94 103(1)mol dm3(1)
=7.96 103if [H3O+] not rounded 4
ALLOW consequential marking on numerical errors
Correct answer with units and some working (4)
(c) (i) Assign the terms acid/base(1)link the pairs(1) 2
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(ii) presence of H+from the first dissociation keeps equilibrium to left
/suppresses ionisation 1
(d) Bromocresol green(1)
pKin/range/colour change (of indicator) lies in vertical section
ORFor alternative indicators pKin/range/colour change (of indicator) lie
outside vertical section(1) 2[12]
27. (a) pentyl dichloroethanoate (1)
ALLOW1,1 OR 2,2-
ALLOW pent-1-yl /all one word
NOTpenten
NOTpentanNOTpentanyl
ester (1)
ALLOWesther 2
(b) (i) using a pipette remove a known volume (say 20 cm3) (1)
remove some solution either with a pipette
ORa known volume / 20 cm3
titrate with an alkali (such as sodium hydroxide) (1)
of known concentration (1) dependent on previous mark ie must have mentioned alkali
IGNOREquenching
using a named indicator eg. phenolphthalein/methyl orange (1)
NOTlitmus / universal indicator
Measure pH on its own 1 (out of 4)
But if calculation fully explained from pH can get full marks 4
(ii)[ ]
[ ] [ ](l)HCCOOH(l)CHCl
(l)HCOOCCHClK
1052
1152
c =
State symbols not required 1
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(iii) C5H10 1.7 (1)3.0
7.1= 5.67(5.7) NOT5.66
CHCl2COOC5H11 0.6 (1)3.0
6.0= 2
(1)for moles at eq by 0.3 in both cases 3
(iv) 2Kc=33.1
3.0/6.0 1.7 / 0.3 (1) =
67.533.1
2
= 0.265 (1) dm3mol1/ mol1dm3(1)
NOT dm3
ALLOW 0.27 / 0.26 / 0.264
Penalise 1 SF or 4SF or more SF but only take off 1 mark maximum in
(iii) and (iv) for significant figure errors
ALLOW TE from expression in (ii)
TE using numbers for (iii) full marks possible 3[13]
28. (a) Enthalpy/heat/energy change for one mole of a
compound/substance/ a product (1)
NOT solid/molecule/species/element
Reject heat released or heat required unless both mentioned
to be formed from its elements in their standard states (1)ALLOW normal physical state if linked to standard conditions
Reject natural state / most stable state
standard conditions of 1 atm pressure and a stated temperature (298 K) (1) 3
Reject room temperature and pressure
Reject under standard conditions
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(b) (i) Bonds broken Bonds made
NN (+)945 6NH ()2346 (1)and
3HH
2253)(
1308)(
+
+(1)
H = 945 +1308 2346= 93 sign and value (1)
H= 93
Accept 46.5 (kJ mol1) with working (4)
= 46.5 (kJ mol1)
sign and value q on 3rdmark (1)
2 4
Accept + 46.5 with workingmax (3)Accept +93 with working max (2)
(ii)
( )Enthalpy
N2 + (3)H 2
(2)NH 3
H
OR
93
Accept 46.5
Correct labelled levels (1)
Reject Reactants and Products as labels
H labelled (1)direction of arrow must agree with thermicity
Accept double headed arrow
Diagram marks cq on sign and value of H in (b)(i)
IGNORE activation energy humps 2
(iii) 350500 C (1)
Accept any temperature or range within this range
higher temperature gives higher rate (1)
but a lower yield because reaction is exothermic (1)
Accept favours endothermic reactionmore than exothermic so
lower yield
OR
Lower temperature give higher yield because reaction is exothermic (1)
but rate is slower (1) 3
Accept cq on sign of Hfin (b)(i) or levels in (ii)
Reject lower temp favours exothermic reaction
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(iv) Iron / Fe (1) IGNORE any promoters
no effect on yield (1) 2
(v) temp would have to be much higher for a reasonable rate then
yield would be too low
lower activation energy implies reasonable rate
ORAllows reaction at a lower temp at a reasonable/fast rate giving
a reasonable yield. 1
Accept rate too slow without catalyst ata temp giving a
reasonable yield
Reject to lower activation energy of reaction
(c) (i) advantage
because smaller number of (gaseous) moles/molecules on rhs (1)
IGNORE any reference to change in rate 2
higher (equilibrium) yield/more NH3 in equilibrium
mixture/equilibrium shifts to right (1)
Reject just more ammonia
(ii) disadvantage
OR
cost (of energy) for compressing the gases/cost of pump
OR
Cost of equipment/pressure not justified by higher yield 1
(plant more) expensive because thicker pipes would be needed
Accept stronger or withstand high pressure for thicker
Accept vessel/container/plant /equipment/reaction vessels for
pipes
Reject just more expensive
Reject just thicker pipes etc
Reject apparatus[18]
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29. IGNORE s.f. throughout this question
(a) Acid
Proton or H+donor
Or forms H+or H3O+(1)
Weak
Accept few molecules dissociate
Or incomplete dissociation
Or partial dissociation
dissociates to a small extent
Or ionises to a small extent (1) 2
Reject not fully dissociated
Or not dissociated fully
(b) 2HCOOH(aq) + Na2CO3(aq) 2HCOONa(aq) + CO2(g) + H2O(l)Or
HCOOH(aq) + Na2CO3(aq) HCOONa(aq) + NaHCO3(aq)
Species + balancing (1)
State symbols (1) Consequential on correct species 2
Accept 2HCOONa(aq) + H2CO3(aq)
Accept HCO2H for the acid
Accept HCO2Na or HCOONa+for salt
(c) (i) one acid: HCOOH
Conjugate base: HCOO
1 mark for both
Accept correct acids and conjugate bases in either order
ACCEPT
HCO2H and HCO2
ORO
HCOH
OHC
O
other acid: H3O+
Conjugate base: H2O
1 mark for both 2
Reject H+for H3O+
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(ii) (Ka) =HCOOH][
]OH][HCOO[ 3+
Accept [H+] instead of [H3O+]
Must use square brackets 1
[HCO2
] and [HCO2H]
(iii) [H+]2= Ka [HCOOH]
OR
Ka=HCOOH][
]H[ 2+
OR
[H
+
]
2
= 1.60 10
4
0.100 (1)
[H+] = 100.01060.1 4
= 4.0 103(mol dm3) (1)
pH = 4.8 scores (2) as square root has not been taken
IGNORE sig figs
Max 1 if [H+]2expression incorrect
pH = log10[H+]
pH = 2.40 (1)
Accept any pH value consequential on ] [H
+
], provided pH < 7Reject pH = 2.39 (is a rounding error) so no third mark
pKa= 3.80 (1)
pH =
Alternative method
acid]log[2
1pK
2
1a (1)
pH = 1.90 (0.50)
pH = 2.40 (1) 3
Reject pH = 2.39 (is a rounding error) so no third mark
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(d) (i) [H+] = Ka[salt]
[acid]
OR
[H+] = 1.60 104
200.0
0500.0(1)
= 4.00 105(mol dm3) (1)
Reject0.400
0.100
pH = 4.40 (1)IGNORE sig figs
Reject 4.39 (rounding error) so no third mark
OR
pH = pKa log10
][HCOO
HCOOH][(1)
pH = log10(1.60 104) log10
200.0
0500.0(1)
Reject0.400
0.100
pH = 3.80 (0.60)
pH = 4.40 (1) IGNORE sig figs 3
Reject 4.39 (rounding error) so no third mark
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(ii) Addition of H+ions:
Accept if described in terms of HA
HCOO+ H+HCOOH(1)
H++ Ashifting to left
Addition of OH
ions
If the ionisation of sodium methanoate shown with
:HCOOH + OHHCOO
+ H2O(1)
then max (1) out of 2 for above equations
Addition of OHions:
Accept molecular equations or equations described in words
or notation involving HA, H+and A.
H++ OHH2Omust befollowed by more dissociation of
HCOOH (to restore [H+])
(buffer solution has) high concentrations
Or a large reservoir of bothHCOOH andHCOO
relative to addedH+/ OH(1)
(hence virtually no change in [H+]) 3
Accept just large reservoir of both HCOOH and HCOO[16]
30. (a) IGNORE s.f. throughout this question
(i) moles SO2(10.0 9.00) = 1.00 (mol)
moles O2(5.00 4.50) = 0.500 (mol)
moles SO39.00 (mol)
all 3 correct (2)2 correct (1) 2
Reject multiples of the stated moles
(ii) All three total number of moles (1)
i.e.
212or)0952.0(
5.1000.1
2==SOX
211or)0476.0(
5.10
500.02
==OX
76
2118 oror)857.0(
5.10
00.93
=SOX
Reject rounding to 1 sig fig
Mark consequential on (a)(i) 1
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(iii) All three total pressure (1)
i.e.
pSO2 = 214or00.2
5.10
00.1
= 0.190 (atm)
pO2 = 212or00.2
5.10
500.0
= 0.0952 (atm)
pSO3 = 712
2136 oror00.2
5.10
00.9
= 1.71 (atm)
Mark consequential on (a)(ii) 1
(iv))0952.0()190.0(
)71.1( 2
2
=pK
Kp= 851 (1)atm1(1)
Mark consequential on (a)(iii) and (a)(iv) 2
Accept answer with units and no working (2)
Accept correct answers between 845 and 855 as this covers
rounding up etc
Reject wrong units e.g. mol1dm3
(b) (i) (Kp) decreases 1
(ii) (Kpdecreases so)
Reject any Le Chatelier argument (this prevents access to 1st
mark)
fraction/quotient22
2
32
pOSOp
SOp
has to decrease (to equal new kp) (1)
so shifts to left hand side (1) this mark onlyavailable if (b)(i) answer was kpdecreases.
Reject shifts to right, even if answer to (b)(i) was kpincreases
(as3SO
p decreases whereas2SO
p and2O
p increase) 2
(c) (i) No effect/none/zero (effect) 1
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(ii) Increases
OR
more SO3/more sulphur trioxide
OR
increases amount of SO3/sulphur trioxide 1
(d) (i) No effect/none/zero (effect) 1
(ii) No effect/none/zero (effect) 1[13]
31. (a) H+ 1
Accept H3O+
(b) HCOOH/HCO2H (1)
HNO3(1)
1 for each extra incorrect answer 2
Accept C and E[3]
32. (a) (i) pH = 3.5 (1)log10[H
+] = 3.5
[H+] = 3.16 104(mol dm3) (1)
2.5(1) 104(mol dm3) based on pH = 3.6 (2 marks) 2
Accept T.E. from wrong pH providing < 7
Accept 3.2 104(mol dm3)
3 104(mol dm3) allowed if evidence of rounding being
applied
(ii) Ka=H]COCHCHCH[
]COCHCHCH][H[
2223
2223
+
(1) 1
Accept version with [H3O+]
Accept molecular formulae
Accept]OHC[
]OHC][H[
284
274
+
(iii) Ka =COOH]CHCH[CH
]H[
223
2+
(1)
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=00660.0
)1016.3( 24(1stmark can be scored here)
= 1.5 105(mol dm3) (1)
Ignore units
Only 2 sig. fig. allowed 2
Accept TE from (i)
Allow any number of s.f. provided consistent with calculation
Reject TE from (ii)
(b) (i) CH3CH2CH2CO2H ((aq)) + NH3((aq))
CH3CH2CH2CO2()NH4(
+) ((aq))
Molecular formulae acceptable 1
Accept eqn via NH4OH CH3CH2CH2CO2+NH4
+
Reject any amide product
(ii) Ammonium butanoate (1)
(Excess) butanoic acid (1)
no TE from (b)(i) 2
Accept ammonium ions and butanoate ions (1)
Reject butanoate ions alone
Reject formulae
(iii) A buffer (mixture) (1)There is a relatively small rise /change in pH (as
aqueous ammonia is added) OWTTE (1)
Mark independently 2
Reject sharp neutralisation point/no change in pH
(iv) There is no large increase in pH / vertical shape to the
graph (at the end-point) OWTTE 1
Accept no sudden change in pH
Reject no indicator has the required pH range
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(v) EITHER
End-point = 30 cm3(1)
[NH3] = (10/30) 0.00660 = 0.00220 (mol dm3) (1)
OR
10 cm3
of butanoic acid contain 6.60 105
molFrom equation this requires 6.60 105mol NH3
From graph, end-point = 30 cm3(1)
[NH3] = 6.60 105 (1000/30)
= 2.20 103/ 0.00220 (mol dm3)(1)
Allow internal TE for 2ndmark based on an incorrect
equivalence point i.e. 0.0660 (mol dm3) 2
Allow T.E. from (b)(i)[13]
33. (a)
A
B
14
12
10
8
6
4
2
0
pH
0 10 20 30 40
Volume 0.1 NH (aq) added/cmM 33
1
Do not worry about general shape of the curve, the scoring points are:
Starting pH ~ 1 and finishing pH between 9 and 11 (1)
Vertical at 25 cm3(1)
Vertical range: at least three pH units in the range 3 to 8e.g. pH range 3 to 6 OR 3 to 7 OR 3 to 8 OR 4 to 7
OR 4 to 8 OR 5 to 8 (1)
(do not need to start/finish on whole numbers)
Accept pH range 3 to 5
Middle of vertical pH range between 4 and 6 (1) 4
(b) Bromocresol green
Indicator(s) CQ on graph [check table on question paper] 1
Accept more than one indicator for extended vertical regions
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(c) pH change around equivalence point too small
OR pH changes over too big a volume (1)
Accept too small a vertical (region)
OR no vertical (region)
OR no point of inflexionOR no sudden change in pH
OR no straight section
for a sharp colour change of indicator (1)
Accept no sharp/clear/precise end point
OR very small range over which indicator changes colour
Reject no suitable indicator
OR No easy colour change
[If say ammonia is a strong base or ethanoic acid is a
strong acid, or both, (0 out of 2)] 2
[7]
34. (a)
H C
O
O C
H
H
C H
H
H 1
(b) ester 1
(c) (i) Moles: C2H5OH: 3.75 (1)
Moles: HCOOC2H5: 2.50 and moles H2O : 2.50 (1) for both 2
(ii) OH]H[HCOOH][C
]OH][HHCOOC[
52
252
=cK 1
Reject obviously round brackets ( )
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(iii)
485.075.3
485.050.0
485.050.2
485.050.2
=cK (1)
Must have clearly divided moles of each component by
0.485 for 1st
mark e.g.[HCOOC2H5] = [H2O] = 5.16 (mol dm
3)
and [HCOOH] = 1.03 (mol dm3)
and [C2H5OH] = 7.73 (mol dm3)
= 3.33 (1) stand alone mark
IGNORE sig.figs. 2
Accept3.750.50
(2.50)K
2
c
= = 3.33 only scores (2) if it is stated
that V cancels either here or in (iv)
If [H2O] omitted in (ii), then answer
Kc= 0.647 mol1dm3
(2) but this will give
Kc= 1.33 mol1dm3with V omitted from calculation (1)
Reject 1stmark if 485 used as V in expression
(iv) No, (as) equal numbers of moles on both sides
OR volumes cancel
OR mol dm3cancel
OR units cancelOR crossing out units to show they cancel 1
Accept equal powers/moles on both sides
OR powers cancel
Mark CQ on Kcexpression in (ii)
Rejectconcentrations cancel
(d) (i) (as reaction) endothermic (1)
Accept exothermic in backward direction (or words to that
effect)
Kcdecreases (1)
If state exothermic in forward direction, 1 mark only (out of 4)for CQ increase in Kc
numerator in quotient (has to) decrease
OR denominator in quotient (has to) increase
OR fraction (has to) decrease (1)
yield of HCOOC2H5decreases (1) 4
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(ii) no effect as catalysts do not affect (the value of) K
OR
no effect as catalysts do not affect the position of equilibrium
OR
no effect as catalysts do not affect the yield
ORNo effect as catalysts increase the rate of the forward and
backward reactions equally/to the same extent
OR
no effect as catalysts only increase the rate
OR
no effect as catalysts only alter the rate
no effect can be stated or implied
IGNORE any references to activation energy 1
Reject just catalysts increase rate[13]
35. (a) The amount of a solidpresent is immaterial since Kc does not depend on
this
OR solids do not appear in expression for equilibrium constants
IGNORE any references to solid in excess. 1
(b) Ag++ IAgI
IGNORE state symbols 1
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(c) Correct answer with some working and correct units scores full marks.
Otherwise steps in calculation must make it reasonably clear to examiner
what is being calculated (QWC).
(initial amount) I= 0.100 mol dm3 0.050 dm3= 5 103mol (1)
Amount Ag+= 0.100 mol dm3 0.031 dm3= 3.1 103mol (1)
equilibrium amount I= 3.1 103mol (1)
Ireacted = (5 3.1) 103mol = 1.9 103mol (1)
If this subtraction is not carried out then the next mark (for calculating
amount of sulphate) can not be awarded.
Thus amount of sulphate = 1.9 103(= 9.5 104mol) (1)
conc iodide =3
3
dm05.0
mol101.3 (= 0.062 mol dm3)
AND conc sulphate = 3
3
dm05.0mol1095.0
(= 0.019 mol dm3) (1)
The mark is for the process of dividing by 0.05 dm3
Kc= 0.019/0.0622= 4.94 (1) Answer must be to 2 or more S.F.
Value consequential on dividing their moles by a volume.
mol1dm3(1) Stand alone 8[10]
36. (a) (i) Pairs: acid NH4+/ammonium ion and base NH3/ammonia
acid H3O+/ hydronium ion and base H2O / water 1
Accept hydroxonium ion
(ii) Ka=]NH[
]OH][NH[
4
33
+
+
ignore lower case k 1
Accept Ka=][NH
]][H[NH
4
3
+
+
Reject answers including [H2O]
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(iii) [H3O+] = 105mol dm3(1)
Assumption ionization of NH4+(negligibly) small (1)
Assumption [NH3] = [H3O+] (1)
Accept [NH4+] = [NH4Cl] or NH4Cl totally ionized
thus [NH4Cl] = (1 105)2/ 5.62 1010
= 0.178 mol dm3(1)
Answer to 2 or more S.F. 4
(iv) QWC
methyl red (1)
indicator constant or pKInmust be near the endpoint pH
OR indicator constant or pKInmust be near 5 (1)2ndmark conditional on correct indicator 2
Accept pKInin the steep part of the graph or it is a weak base-
strong acid titration
(b) CN+ H2O HCN + OH
IGNORE state symbols 1
Accept instead of
(c) (i) nucleophilic addition 1
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(ii)
(:)CN
C O CNC
O(:)
(1) for both arrows (1) for intermediate
(1)
C CN
O(:) H CN
C
OH
CN + (:)CN
OR for second step
C
O(:)
CN
H
C
OH
CN
(1) 3
Fish hook arrows (penalise once)
Ignore the groups attached to the carbonyl carbon throughout
The intermediate is not consequential on their first step
The minus of the cyanide ion can be on either the C or the N
The arrow can start from the minus of
CN in step 1 (but notfrom the minus of CN) and can start from the minus of Oin step 2
The arrow from the bond must not go past the O atom
Lone pairs not essential
Single step addition of HCN or initial attack by H+/HCN scores zero
Autoionisation of C=O can only score the last two marks ie max 2
(iii) QWC
if too acidic too small a concentration of cyanide ions (1)Accept not enough / too little CN
if too alkaline too little HCN to donate the proton in the last step
OR H+ion concentration too low (1) 2
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(d) (i) rate = k[CH3CH2CH2Cl] [CN]
Must be an equation
Must be [ ] NOT ( )
Ignore upper case K 1
Accept R or r for rate C3H7Cl] / [1-chloropropane]/[chloropropane]
Accept [cyanide ion]/[cyanide]
Reject [KCN]
(ii)
Curly arrow (1)
Curly arrow (1) Transition state (1)
H
CH Cl
C HNC
2 5
H H
C ClNC
C H2 5
H
CNC H + Cl
C H2 5
Must have partial bonds in transition state
CN and Cl must be on opposite sides of central C in the
transition state
Accept negative charge on N of cyanide ion 3
Mechanism based on SN1 scores 0
Reject fish hook arrows (penalise once)
Reject arrow from N of CN[19]
37. (a) (i) [6 188.7 + 4 210.7] [4 192.3 + 5 205] (1)
+180.8 J mol1K1(1)
Accept +181 J mol1K1
Reject internal TE
1 for missing + sign/missing or incorrect units but penalise only
once in part (a)
[IGNORE sig fig] 2
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(ii) yes, as 9 molecules of gas are being changed to 10 molecules of
gas (therefore increase in disorder) 1
Accept TE from (i)
Not just 9 molecules going to 10 molecules
(iii) 905.6 1000 /1123 (1)
+ 806.4 J mol1K1/ 0.8064 kJ mol1K1(1)
[IGNORE SF] 2
Accept + 806 J mol K1
(iv) +987.2 J mol1K1 1
Accept +987 J mol1K1
allow TE from (i) & (iii)
No TE if J mol1K1added to kJ mol1K1
(v) All products/reaction goes to completion because Stot> 200 J
mol1K1/Stotis very large[Needs to be consistent with (iv)] 1
(vi) catalysed pathway should have lower Eathan uncatalysed pathway
and the peak of the curve should be above the energy level of thereactants (1)
Energy of products should be lower than energy of reactants (1) 2
(b) (i)
22
2
2
pOpNO
pNOKp
= 1
Accept
22
22
pONOp
NOp
Reject [ ]
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(ii) mole fraction NO2=5
95.4or 0.99 (1)
mole fraction NO =5
025.0or 0.005
OR
mole fraction O2=5
025.0or 0.005 (1)
Kp= 33
22
)15.1()005.0(
)5.1()99.0(= 5227200 / 5.2 106(1)
atm1(1) unit mark independent but consistent with
expression used in calculation.
IGNORE SF 4
Correct answer for Kpalone = 3 max
(iii) Equilibrium lies to RHS/products side as Kpis large
OR reaction wont go to completion as Kp< 1010
Must be consistent with (ii) 1
Allow TE from b(ii)
(iv) Kpremains unchanged as pressure does not affect it / only
temperature affects Kp
(1)
partial pressure of NO2 increases as eqm moves to side of
fewest (gas) molecules/RHS (1)
or
Partial pressure of NO2 increases as pp = mole fraction total
pressure 2
Accept justification in terms of entropy[17]
38. (a) (i)H]ClCOCH[
]H][ClCOCH[
22
22+
=aK 1
Accept [H3O+] in place of [H+]
allow one set of sq brackets to be missing
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(ii) [H+]2= 1.3 103 0.001 (1)
= 1.3 106
[H+] = 1.3 106
1.14 103(1)
pH = log 1.14 103= 2.9(4) (1)
[IGNORE SF] 3
(iii) Trichloroethanoic, as it has the largest Kavalue (1)
and has (3 electron withdrawing) chlorine atoms to stabilise
the anion formed (on dissociation). (1) 2
(b) (i)
H C
H
Cl
O
C O C
H
H
H
ester group (1)
rest of molecule (1) dependent on first mark
(must be fully displayed)
methyl chloroethanoate (1) 3
No transferred error for name
(ii) ester(s) 1
Reject ether
(iii) nucleophile, (1)
as it has a lone pair (of electrons) on the (hydroxyl) oxygen (1)
which can attack the positive carbonyl carbon on the acid (1) 3
2ndand 3rdmarks could be obtained by use of a diagram
Reject attack by CH3O
(iv) (reflux) heat with NaOH(aq) (1)
(cool) and add HCl(aq) (1)
OR
reflux (1) [must be in context]
with HCl (1) 2[15]
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39. (a) N/N2goes from 0 to 3 = reduction (1)
H/H2goes from 0 to (+)1 = oxidation (1) 2
If the oxidation number of N goes down hence reduced and the
oxidation number of H goes up and hence oxidised (max 1)
If all O.N. correct but fails to state which is oxidation and
which is reduction scores 1.
If all O.N. correct but both reactions misclassified, scores zero.
Any answer not referring to nitrogen or hydrogen scores zero.
(b) (i) Calculation of bonds broken 463 3 + 944/ (= 2252) (1)
Calculation of bonds made 3886/ (= 2328) (1)
H = 76 (kJ mol1) (1)mark consequential on numerical values calculated above 3
Correct answer with some working scores 3 marksCorrect answer alone scores 2 marks
(ii) Average / mean bond enthalpy used for NH bond / ammonia 1
Reject just average bond enthalpies used
(iii) Thermodynamic
Accept H negative / reaction exothermic
:
energy level of products lower than that of reactants
OR
energy released in bond formation > energy used to break bonds (1)
kinetic
because strong NN (1)[confusion between thermodynamic and kinetic loses first 2 marks]. 3
:
high activation energy (1)
Accept because NN is 944/ total bond breaking energy is
high/2252(kJ mol1)
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(c) (i) QWC
One way
Accept moving faster
temperature increase therefore molecules have greater (average
kinetic) energy (1)
more molecules/collisions have E Eact(1)
Therefore a greater proportion of/ more of the collisions are
successful (1)
Ignore greater frequency of collision
Accept E > Eactparticles for molecules
greater frequency of successful collisions/ more successful
conditions per unit time
Reject just more successful collisions
Another way
Accept platinum catalyst
addition of (iron) catalyst (1)
Reject incorrect catalyst
provides alternative route of lower activation energy (1)
EITHER:
A greater proportion of /more of the molecules/collisions have E Ecat/a greater proportion of collisions are successful
Reject just more successful collisions
OR provides (active) sites (where reactant molecules canbond / be adsorbed) (1)
Ignore any answers referring to pressure or concentration.
Do not penalise just more collisions are successful more
than once 6
(ii) QWC
Decrease temperature (1)
because (forward) reaction exothermic (1)
increase pressure (1)
because more moles (of gas) on left (1) 4
Accept low temperature H is negative
Answer based on endothermic reaction scores 0
Accept high pressure
Accept molecules for moles[19]
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40. (a) (i) To slow down the reaction/to stop the reaction
OR to quench the reaction
OR to freeze the (position of) equilibrium OWTTE (1)
so that the (equilibrium) concentrations/amounts do not change (1) 2
Accept to stop equilibrium shifting to the left
(ii) First mark:
][][)()( 22 gg
IH =
OR
Use of (5.0 104)2(1)
If [HI] not squared, first mark only.
Second mark:
019.0
)100.5(][
242
)(
=gHI
OR
0.019 =2
(g)
24
]HI[
)100.5(
OR
[HI(g)] =
019.0
)100.5( 24(1)
Third mark:
[HI(g)] = 3.6 103(mol dm3) (1)
Correct answer scores 3 marks.
Ignore state symbols.
Ignore units unless wrong.
Ignore s.f. 3
If first mark not awarded, total (0)
(b) (i)
22
2
IH
HIp
pp
pK
=
Ignore position of any ( ) 1
[ ] scores (0)
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(ii) Each step of this calculation must be looked at.
1stmark is for calculating equilibrium moles2= 0.22= 0.2
= 1.6 (1)Mark consequentially
2ndmark is for dividing these by 2 (to get mole fractions)
1.00.2
2.02 ==Hx
1.00.2
2.02
==Ix
8.00.2
6.1==HIx (1)
Mark consequentially
3rdmark is for multiplying by 1.1 (to get partial pressures)
1.10.2
2.0P
2=H
= 0.11 (atm)
110.2
2.0P
2=I
= 0.11 (atm)
110.2
6.1P =HI
= 0.88 (atm) (1)
Mark consequentially
4thmark is for substituting into their expression and calculating Kp
)11.0()11.0(
)88.0( 2
=pK
= 64 (1)
Ignore s.f.
Correct answer with no working scores (1) 4
If moles HI given as 0.8, Kp= 16 max (3)
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(iii) Same number of moles on each side
OR
(Total) pressure cancels
OR
(Pressure) units cancel
(May be shown by crossing out etc. in b(ii)) 1
Accept Powers cancel
OR
They cancel
OR
Same number of molecules on each side
Reject Partialpressures cancel
OR
mol dm3cancel[11]
41. (a) (i) One acid: CH3CH2COOH(aq)
Conjugate base: CH3CH2COO(aq) (1)
Other acid: H3O+(aq)
Conjugate base: H2O(l) (1)
Ignore state symbols 2
Accept correct acids with conjugate bases in either order
(ii) WEAK: dissociates/ionises to a small extent (1)OWTTE
Accept Few molecules dissociate
Accept Incomplete or partial dissociation
Accept Does not fully dissociate
Reject ions partially dissociate
ACID: proton donor (1) 2
Accept Produces H3O+/ hydrogen / H+ions
Reject just contains H3O+.
(b) (i)][
]][[
23
323
COOHCHCH
OHCOOCHCHKa
+
= 1
Accept [H+] instead of [H3O+]
Reject any expression containing [H2O]
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(ii) ([H+] =) 3.63 104(mol dm3) (1)
Or 103.44
If Kaexpression incorrect in (b)(i) or [H+] not squared, only 1st
mark available
[CH3CH2COOH] = 5
2
1030.1
]H[
+
Or
[CH3CH2COOH] = 5
24
1030.1
)1063.3(
(1)
= 0.010 (1)(mol dm3) (1)
ASSUMPTIONS
First assumption mark:
negligible [H+] from ionisation of water Or [CH3CH2COO] = [H+] (1)
:
Accept No other source of H+ions
RejectJust CH3CH2COO= H+ (ie no square brackets)
Second assumption mark:
Ionisation of the (weak) acid is negligible
Or x[H+] x where x is initial concentration of CH3CH2COOH
Or [H+]
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(ii) Working MUST be checked
First mark:
[H+] = Ka[base]
acid][(1)
Accept Ka=acid][
salt][]H[ +
Second mark:
Correct [acid] = 0.0025 and [salt] = 0.00375 (1)
Third mark:
Calculation of pH correct consequential on [acid] and [salt] used.
[H+] = 1.30 10500375.0
0025.0
= 8.67 106(mol dm3)
pH = 5.06 (1)
Accept if [salt] and [acid] inverted, pH is 4.71 (2 marks)
Accept inverted with the original concentrations, pH = 5.19 (1
mark)
Ignore sig fig
OR
First mark:
pH = pKa
[salt]
acid][log10 (1)
Reject in both cases, if [acid] = [0.0100] and [salt] =
[0.00500], pH = 4.59 (2 marks)
Second mark:
Correct [acid] = 0.0025 and [salt] = 0.00375 (1)
Third mark:
Calculation of pH correct consequential on [acid] and [salt] used.
pH = 4.89 ]00375.0[
]0025.0[log10 (1)
= 4.89 (0.18)
= 5.07 (1)
Accept 5.06
Ignore sig fig 3[15]
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42. (a) (i)
22
2
p
)(K
ON
NO
pp
p
=
Allow answer with brackets and/or x omitted
Ignore (g) and eq 1
Accept
22
2
pKON
NO
pp
p
=
Reject anything in [ ]
(ii) Same number of moles on each side of the equation OR
The (partial pressure) units all cancel out (in the expression for Kp) 1
(b) (i) (pNO)2
= 0.87 0.23 5.0 1031
(1)= 1.0 1031
pNO= (1.0 1031)
= 3.2 1016(atm) (1)
Accept 3.16 1016(atm) (1)
Ignore sig fig
Mark consequentially only if based on reciprocal of
correct expression in (a)(i) 2
(ii) 0.87 + 0.23 (+ 3.2 1016) = 1.10 / 1.1 (atm)
Allow TE from (b)(i) 1
Reject answer based on adding2 pNO
(iii) pNOdoubles/will become 6.4 1016atm (1)
Kpremains constant/is (still) 5.0 1031(1)
Ignore any neutral qualifications to these answers 2
Accept pNOwill increase
Reject more than double
Reject answers with incorrect reasoning
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(c) (i) Reaction will occur, but (very) little
NO is formed
OR
the equilibrium mixture is mainly
(unreacted) N2and O2 1
Accept reaction occurs, but equilibrium lies (very much) to the
left
Reject Reaction is more likely to occur from right to left OR
Reverse reaction is favoured, unless included with
acceptable answer
(ii) No change of state of any of the components is involved
(as the gases are heated up) OWTTE
OR
All components are gases (at these temperatures)IGNORE Any reference to the number of particles involved 1
(iii) (H is positive so)T
H = Ssurroundingswill be negative
No mark for negative alone 1
Accept negative, since for an endothermic reaction energy is
taken from the surroundings causing a decrease in disorder /
reduction in entropy
(iv) (As T increases) Ssurroundingsbecomes greater/less negative/more positive, so Stotal(also) becomes greater/lessnegative/more positive/increases 1
Accept Ssurroundingsbecomes smaller, if qualified
, e.g.
becomes closer to zero
(d) Equilibrium might not have been reached (in the very short time the
gases are present in the engine)
Ignore references to the fact that the system is not closed 1
Accept other gases are present in the air (apart from N2and
O2)
Accept temperature inside engine may be less
Accept actual (total) pressure may be less than that assumed
than 1500K
[12]
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43. [C6H5CO2H] = (1/5 0.010 =) 0.002(0) mol dm3
[C6H5CO2] = (4/5 0.020 =) 0.016 mol dm3
Accept pKa= 4.20 (1)
Both correct (1)
Accept0.016
0.002(1)
[H+] (= Ka [C6H5CO2H] / [C6H5CO2]
= 6.3 105 0.0020 / 0.016)
= 7.875 106(1)
Accept pH = (4.20 + 0.90) = 5.1 / 5.10 (1)
Do not penalise SF for the first two marks
pH = log[H
+
] = 5.1 / 5.10 (1)Mark for final answer must be dependant on valid working
e.g. correct [acid]/[base] ratio.
Correct answer with no working (1)
Allow internal TE
e.g. an [acid]/[base]ratio of0.010/0.020leads to a pH of 4.50 (2)
Reject 5.104 or 5
Reject 1 or >3 sig. fig.[3]
44. (a)
42
2
2
ON
NO
pp
pK =
IGNORE UNITS HERE 1
Reject [ ]
(b) (i)2NO
p = 0.8 1.1
= 0.88 (atm)
and
22ONp = 0.2 1.1
= 0.22 (atm) (1)
Kp=)22.0(
)88.0( 2
Kp= 3.52 (1)
atm (1)
SECOND MARK IS CQ ON PARTIAL PRESSURES
ASCALCULATED 3
(ii) First mark:
42ONX = 0.10
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2NOX = 0.90 (1)
Reject B
Second mark:
Kpconstant or
use of Kp= 3.52 or
use of Kpcalculated in (b)(i) (1)
Third mark:
Value of PTwith some working e.g.
3.52 =T
2T
P
)P(
42
2
ON
NO
X
X
3.52 =10.0
81.0 PT
PT= 0.435 (atm) (1)
Mark CQ on first and second answers to (b)(ii)
Accept in range 0.43 to 0.44
THIRD MARK NOT AVAILABLE IF KpEXPRESSION
DOES NOT CONTAIN A p2TERM 3
(c) (i) Increases / gets larger/ gets bigger/ goes up/greater 1
Reject more
(ii) First mark:
Fraction/quotient/
42
2
2
ON
NO
p
p/numerator has to increase
(to equal new Kp) (1)
Second mark (can only be awarded for an answer that refers
to the fraction/quotient above):
EITHER
so shifts to RIGHT hand side (as2NO
p and42ON
p ) /
goes in forward direction (as2NO
p and42ON
p )
OR
so (more) N2O4changes to NO2
OR
so (equilibrium) yield of NO2increases (1) 2
Mark consequentially on decreases in (i)
Le Chatelier argumentscores (0)[10]
45. (a) (i) H2O(l) H+(aq) + OH(aq)
OR
2H2O(l) H3O+(aq) + OH(aq)
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IGNORE STATE SYMBOLS 1
Reject if afull arrow is shown in the equation
(ii) KW= [H+
(aq)][OH
(aq)]
OR
KW= [H3O+
(aq)][OH
(aq)]
IGNORE STATE SYMBOLS 1
If [H2O] included (0).
Reject Kw= [H+]2
(iii) pH = log10[H+]
OR
pH = log10[H3O+]
OR
in words 1
Accept pH = lg 1/[H+]
(iv) KW= [H+][OH]
5.48 1014= [H+]2(1)
[H+] = 141048.5
[H+] = 2.34 107(mol dm3)
pH = 6.6(3) (1)
correct answer with no working (2) 2
pH = 13.3/13.6 scores (0)
(v) (In pure water)
[H+] = [OH]
OR
equal concentrations of H+and OH 1
(b) (i) 12.5 1
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(ii) 4.8 / 4.9
[no consequential marking from (i)] 1
Reject 5 or 5.0
(iii)][
]][[
3
3
COOHCH
HCOOCHKa
+
=
OR
][
]][[
3
33
COOHCH
OHCOOCHKa
+
= 1
Reject expressions containing [H2O]
OR
Reject HA
generic equations
(iv) (at half-neutralised point so)
pKa= 4.8
Mark CQ on (ii)
RejectjustpH = 4.8 as already credited in (b)(ii)
OR
pH = pKa(1)
Ka= antilog10(4.8)
Ka= 1.6 105(mol dm3) (1)
Mark CQ on pKa
Accept if pKa = 4.9, Ka = 1.3 105
Reject answers to other than 2 s.f.
Must be to two sig figs
CORRECT ANSWER WITH OR WITHOUT WORKING(2) 2
Reject 2.5 109
scores (0)
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(c) Phenolphthalein:
changes colour (OWTTE) in vertical part of the graph
OR
changes colour within a stated range anywhere from 7 to 11 (1)
Reject if colour change pink to colourlessMethyl orange
changes colour at a low(er) pH
OR
has already changed colour
OR
changes colour before the vertical (section) (1)
Allow range for methyl orange of 3 to 6 or colour change takes
place below pH = 7
Rejectjust methyl orange changes colour outside the vertical
range
[NB There must be a statement about methyl orange for second mark] 2[13]
46. (a) (i) Liquids are more disordered than solids/ solids are more
ordered than liquids/ solids are less disordered than liquids /
liquids are less ordered than solids 1
Accept more ways of arranging energy in a liquid because of
translation/rotation energy
Reject just more ways of arranging energy
(ii) (165 + 217.1 166.5 =) + 215.6 OR +216 (J mol1K1)
+ sign essential 1
Accept +(0).2156 kJ mol1K1
OR +0.216 kJ mol1K1
Reject 215 J mol1
K1
Reject 0.215 kJ mol1K1
(iii) Yes because
The products include a gas (1)
Accept solid goes to liquid and gas for first mark
One mole/molecule goes to two moles/molecules (1) 2
1 reactant goes to 2 products does not get 2ndmark
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(b) Ssurroundings=T
H
OR
298
123800
(1)
= 415 J mol1K1(1) 2
Accept 0.415 kJ mol1
K1
Accept 415.4 J mol1K1
Accept final answer with no working (2)
Allow j for J
Reject full calculator display eg 415.4362416
Reject more than 2 dp e.g. 415.436
(c) (i) Stotal= 415 + 216 = 199
or 199.8 or 200) (J mol1K1)
IGNORE 4thsignificant figure 1
Accept 0.199 kJ mol1K1
ALLOW TE from(a)(ii) and (b)
(ii) reactants predominate / equilibrium lies well to the left
OR
Equilibrium completely to the left 1
ALLOW TE from (c)(i)
(d) (i)
5
23
PCl
CLPCl
pp
ppK
= (1)
IGNORE state symbols or lack of them unless (s) or (l)
Units atm (1) 2
Accept capital P
Accept use of ( )
If expression the wrong way up allow second mark if units given
as atm1
Reject use of [ ]
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(ii)
Substance Moles at
start
Moles at
equilibrium
Peq/atm
PCl5(g) 0.20(1) 0.25
0.15
4.32
= 2.592
PCl3(g) 0.05
0.25
0.05 4.32
= 0.864
Cl2(g) 0.05 0.864
Total number of moles
at equilibrium
0.25
All three(1) All three(1)
Allow consequential marking across columns 3
If moles at eqm are given as 0.025 for PCl3(g) and for Cl2(g)
then 4thcolumn should be 3.24, 0.54 and 0.54
and gets 2 (out of 3)
(iii) ()592.2(
864.0864.0 =pK )
= 0.288 (atm) 1
ALLOW TE from di and from dii
Common wrong values above gives 0.090
ALLOW 0.29
Reject 0.3
reject 0.28
(iv) ANo change because Kpdepends only on temperature /
number of moles would change in same proportion (1)
B Increase because reaction is endothermic (1)
OR
entropy arguments 2
If both changes correct but no explanations then 1 (out of 2)[16]
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47. (a) (i) Ka=]CO[
]HCO][H[
2
3
+
(1)
mol dm3(1)
If H2O is included as denominator then allow only the 2nd
mark if no units suggested 2
(ii) pKa = log Ka / lg Ka / log10Ka 1
Accept Ka= 10pKa
(b) A solution which does not change its pH value (significantly) (1)
May be shown using an equation
When some/small amount of acid or alkali is added (1) 2
(c) Acting as a base because it is accepting
a proton (to form H2CO3/CO2+ H2O) 1
(d) (i) Before race 7.4 = 6.5 log[base]
acid][
Log[base]
acid][= 0.9 (1)
[base]
acid][= 0.126 (1) 2
Accept 0.13
Reject 0.12
(ii) Before race
[CO2] = 0.126 0.0224 = 2.82 103
OR
2.52 102 2.24 102= 2.8 103 1
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(iii) Hypothesis I would result in an increase in
[CO2] / [HCO3] / [CO2+ HCO3
]
OR
Hypothesis II would produce greater acidity without
additional [CO2] / [HCO3
] / [CO2+ HCO3
] (1)
The table shows a fall in [CO2] / [HCO3] / [CO2+ HCO3
]
and therefore Hypothesis II must be favoured. (1) 2[11]
48. C[1]
49. C[1]
50. A[1]
51. C[1]
52. (a) C 1
(b) D 1
(c) B 1[3]
53. (a) A 1
(b) D 1
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(c) C 1[3]
54. (a) methyl butanoate
Accept Methyl butaneoate 1
Reject an missing
(b) the other three substances can form
intermolecular hydrogen bonds with themselves but the ester cannot. 1
Reject Discussion of London Forces
(c) Hydrolysis 1
(d) QWC
Must cover advantages and disadvantages. Must notbe contradictory
Advantages to manufacturers: (any two)
not dependent on weather, seasons etc
consistent taste /concentration/more consistent
quality
or alternative ideasDisadvantages to consumers : (any two)
some people put off by non-natural food
may not taste the same as natural product which may contain
other impurities
unable to describe the product as organic
or alternative ideas 4
Reject cost with no justification
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(e) Kc =O(l)](l)][HCOOCHHC[
OH(l)]HCOOH(l)][CHC[
2373
373 (1)
Accept eq subscripts
Moles at Concentration /
equilibrium mol dm3
butanoic acid= 4.4/88 = 0.05 1.67
methanol 0.05 1.67
ester (methyl butanoate) 0.05 1.67
water 0.95 31.7
all four equilibrium moles = (1)
Conc at equilibrium = equilibrium moles 0.030 (1)
Kc =7.3167.1
67.167.1
(1)= 0.053 (1)
ignore significant figures unless value given to 1 s.f.
The units cancel because both the top and bottom of the fraction
have units of concentration squared.
Or same number of moles on both sides of the equation (1) 5
Reject absence of square brackets[12]
55. (a) Value of equilibrium constant increases (1) 1
(b) QWC
If the equilibrium constant increases then more products will be formed (1)
And the position of equilibrium will move to the right (1) 2[3]
56. (a) Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
Accept state symbols omitted 1
(b) (i) Positive because a gas is given off (1)
which is more disordered and so has more entropy (1) 2
(ii) Positive because the reaction is exothermic (1)
and = H/T (1) 2
(iii) Positive because the reaction occurs / total entropy change
is the sum of the two positive values above. 1
(c) (i) Surface coated with magnesium oxide (which would react to
form water rather than hydrogen). 1
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(ii) QWC
Initial number of moles of HCl = 20 1 /1000 = 0.02
Number of moles of Mg = 0.1 / 24 = 0.00417 (1)
number of moles of HCl which reacts is 0.00834 (1)
Therefore number of moles of HCl left = 0.01166 (1)Ignore sig figs
so the concentration nearly halves which would significantly
reduce the rate and so make the assumption that the initial
rate is proportional to 1/time invalid / inaccurate. (1)
Increase the volume of acid to (at least) 50 cm3(1)
Or measure the time to produce less than the full amount of gas
Or use a smaller piece of magnesium. (1) 5
(iii) Energy given out = 467 000 0.1/24 J = 1 946 J20 4.18 T = 1 946 (1)
T = 23.3(o)(1)
Accept units of degrees celsius or Kelvin
This temperature change would significantly increase
the rate of the reaction (1)
Carry out the reaction in a water bath of constant
temperature/use a larger volume of more dilute acid (1) 4
(iv) At 329 time 4s 1/time = 0.25 s1ln(rate) = 1.39 (1)
At 283 time 124s 1/time = 0.00806 s1ln(rate) = 4.82 (1)
[graph to be drawn]
Plot line with new gradient = 3.43 / 0.00049
= 7 000 (1)
Accept 6800 to 7200
Activation energy = + 7 000 8.31
= + 58.2 kJ mol1(1) 4
(v) QWCRate of reaction reduced because less surface area in contact
with the acid. (1) 1
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(vi) Any two
Repeat the experiment at each of the temperatures
obtain an initial rate eg by measuring the volume of gas
given off before the reaction is complete.
Other sensible suggestions. 2
(vii) The rate should be lower, since ethanoic acid is a weaker acid
(compared to hydrochloric acid) and so there will be a lower
concentration of hydrogen ions present. 1[24]
57. QWC
Answer must be given in a logical order, addressing all the points using preciseterminology
Collision frequency increases as particles moving more quickly (1)
More collisions have sufficient energy to overcome activation energy /
more molecules on collision have energy activation energy (1)
A greater proportion of collisions result in reaction (1)
Collision energy has greater effect (1)
Homogeneous all in same phase and heterogeneous in different
phases / gas and solid (1)
No need to separate products from catalyst (1) 6
Reject more collisions
Reject more successful collisions[6]